A cylinder is filled with a liquid of density d upto a height h. If the beaker is at rest ,then the mean pressure on the wall is?​

An action which would help in preventing and coping with heat-related conditions is: A. Drinking water.

Heat can be defined as a form of energy that is transferred from a physical object (body) to another, as a result of a difference in temperature. Also, heat is a condition of weather that is generally characterized by a high degree of temperature.

This ultimately implies that, heat is most likely to cause dehydration and high body temperature.

In order to prevent and cope with heat-related conditions, you should ensure that you drink water at regular intervals for hydration.

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Answer:

616000 J.

Explanation:

Bi. Determination of the work done.

Force (F) = 220 N

Distance (s) = 2800 m

Workdone (Wd) =?

Workdone is simply defined as the product of force and the distance moved in the direction of the force. Mathematically, is can be expressed as:

Workdone = Force × distance

Wd = F × s

With the above formula, we can obtain the Workdone as follow:

Force (F) = 220 N

Distance (s) = 2800 m

Workdone (Wd) =?

Wd = F × s

Wd = 220 × 2800

Wd = 616000 J.

Answer:

Explanation:

Current  has speed vW with respect to the shore and boat has speed vB with respect to water or current so speed of boat  with respect to shore

vW + vB .

Distance travelled with respect to shore by boat = D

time ( tout ) = distance / speed with respect to shore

tOut = D / ( vW + vB )

When the boat travels upstream , its velocity with respect to shore

= ( vB - vW ) , vB must be higher .

tin = D /  ( vB - vW )

3 ) tin = D /  ( vB - vW )

170 = 120 / (vB - 0.3 )

(vB - 0.3 ) = 12 / 17 = .706

vB = 1.006 m / s

4 )

tOut = D / ( vW + vB )

= 120 / ( .3 + 1.006 )

= 92.26 s

Time taken by a body is ratio of the distance traveled by it to the speed.

          [tex]t_{out}=\dfrac{D}{v_B+v_W}[/tex]

           [tex]t_{in}=\dfrac{D}{v_B-v_W}[/tex]

The net speed of the boat is upstream speed. The difference of the speed of the boat is downstream speed.

Given information-

The speed of the boat with respect to shore is [tex]v_w[/tex].

The speed of the boat in downstream with respect to water is [tex]v_B[/tex].

The distance traveled by the boat is [tex]D[/tex] in time [tex]t_{out}[/tex].

Time taken by a body is ratio of the distance traveled by it to the speed.

        [tex]t_{out}=\dfrac{D}{v_B+v_W}[/tex]

        [tex]t_{in}=\dfrac{D}{v_B-v_W}[/tex]

Now the distance is 120 m, the value of [tex]t_{in}[/tex] is 170 s and [tex]v_W[/tex] 0.3 m/s. Thus,

         [tex]170=\dfrac{120}{v_B-0.3}\\v_B-0.3=\dfrac{120}{160} \\v_B=0.706+0.3\\v_B=1.006[/tex]

Hence the speed of the boat with respect to the water is 1.006 m/s.

          [tex]t_{out}=\dfrac{120}{1.006+0.3}\\t{out}=\dfrac{120}{1.306} \\t{out}=91.9[/tex]

Hence the time it takes the boat to move a distance D downstream is 91.9 seconds.

Thus,

          [tex]t_{out}=\dfrac{D}{v_B+v_W}[/tex]

           [tex]t_{in}=\dfrac{D}{v_B-v_W}[/tex]

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Answer:

      I =  [tex]\frac{1}{4\pi \ r^2}[/tex]

we see the intensity decreases with the inverse of the distance squared

Explanation:

Intensity is defined as power per unit area,

           I = P / A

in this case we have that the sound is emitted in a spherical form therefore the area is

           A = 4 pi r2

therefore the intensity is

          I =  [tex]\frac{1}{4\pi \ r^2}[/tex]

as we see the intensity decreases with the inverse of the distance squared

Answer:

1) The total charge of the top plate is 0.008 C

b) The total charge of the bottom plate is -0.008 C

2) The electric field at the point exactly midway between the plates is 0

3) The electric field between plates is approximately 1.1294 × 10¹² N/C

4) The force on an electron in the middle of the two plates is approximately 1.807 × 10⁻⁷ N

Explanation:

The given parameters of the parallel plate capacitor are;

The dimensions of the plates = 4 × 2 cm

The distance between the plates = 10 cm

The surface charge density of the top plate, σ₁ = 10 C/m²

The surface charge density of the bottom plate, σ₂ = -10 C/m²

The surface area, A = 0.04 m × 0.02 m = 0.0008 m²

1) The total charge of the top plate, Q = σ₁ × A = 0.0008 m² × 10 C/m² = 0.008 C

b) The total charge of the bottom plate, Q = σ₂ × A = 0.0008 m² × -10 C/m² = -0.008 C

2) The electrical field at the point exactly midway between the plates is given as follows;

[tex]V_{tot} = V_{q1} + V_{q2}[/tex]

[tex]V_q = \dfrac{k \cdot q}{r}[/tex]

Therefore, we have;

The distance to the midpoint between the two plates = 10 cm/2 = 5 cm = 0.05 m

[tex]V_{tot} = \dfrac{k \cdot q}{0.05} + \dfrac{k \cdot (-q)}{0.05} = \dfrac{k \cdot q}{0.05} - \dfrac{k \cdot q}{0.05} = 0[/tex]

The electric field at the point exactly midway between the plates, [tex]V_{tot}[/tex] = 0

3) The electric field, 'E', between plates is given as follows;

[tex]E =\dfrac{\sigma }{\epsilon_0 } = \dfrac{10 \ C/m^2}{8.854 \times 10^{-12} \ C^2/(N\cdot m^2)} \approx 1.1294 \times 10^{12}\ N/C[/tex]

E ≈ 1.1294 × 10¹² N/C

The electric field between plates, E ≈ 1.1294 × 10¹² N/C

4) The force on an electron in the middle of the two plates

The charge on an electron, e = -1.6 × 10⁻¹⁹ C

The force on an electron in the middle of the two plates, [tex]F_e[/tex] = E × e

∴ [tex]F_e[/tex] = 1.1294 × 10¹² N/C ×  -1.6 × 10⁻¹⁹ C ≈ 1.807 × 10⁻⁷ N

The force on an electron in the middle of the two plates, [tex]F_e[/tex] ≈ 1.807 × 10⁻⁷ N

Explanation:

this is the answer for your question. if you have any doubt.

you can send your doubt to:6369514784(what's app)

Answer:

2.18 kg

Explanation:

The frequency of a wave in a stretched string f = n/2L√(T/μ) where n = harmonic number, L = length of string, T = tension = mg where m = mass of object on string and g = acceleration due to gravity = 9.8 m/s² and μ = linear density of string.

For string 1, its fundamental frequency f  is when n = 1. So,

f = 1/2L√(T/μ) =  1/2L√(mg/μ)

Now for string 1, L = 0.50 m, m = 20 kg and μ = 5 g/m = 0.005 kg/m

substituting the values of the variables into f, we have

f = 1/2L√(mg/μ)

f = 1/2 × 0.50 m√(20 kg × 9.8 m/s²/0.005 kg/m)

f = 1/1 m√(196 kgm/s²/0.005 kg/m)

f = 1/1 m√(39200 m²/s²)

f = 1/1 m × 197.99 m/s

f = 197.99 /s

f = 197.99 Hz

f ≅ 198 Hz

For string 2, at its third harmonic frequency f'  is when n = 3. So,

f' = 3/2L√(T/μ) =  3/2L√(mg/μ)

Now for string 2, L = 0.50 m, m = M kg and μ = 5 g/m = 0.005 kg/m

substituting the values of the variables into f, we have

f' = 3/2L√(Mg/μ)

f' = 3/2 × 0.50 m√(M × 9.8 m/s²/0.005 kg/m)

f' = 3/1 m√(M1960 m²/s²kg)

f' = 3/1 m√M√(1960 m²/s²kg)

f' = 3/1 m √M × 44.27 m/s√kg

f' = 132.81√M/s√kg

f' = 132.81√M Hz/√kg

Since the frequency of the beat heard is 2 Hz,

f - f' = 2 Hz

So, 198 Hz - 132.81√M Hz/√kg = 2 Hz

132.81√M Hz/√kg = 198 Hz - 2 Hz

132.81√M Hz/√kg = 196 Hz

√M Hz/√kg = 196 Hz/138.81 Hz

√M/√kg = 1.476

squaring both sides,

[√M/√kg] = (1.476)²

M/kg = 2.178

M = 2.178 kg

M ≅ 2.18 kg

Answer:

predict annual flooding

Answer:

I'm pretty sure it is b make their calander

Answer:

put it in medium to full heat and set an alarm so that it doesn't burn

Answer:

when you're cooking  your pancakes, use clarified butter and cook on medium heat. 

Answer:

Plant cell Animal cell

2. Have a cell membrane. 2. Have no chloroplasts.

3. Have cytoplasm. 3. Have only small vacuoles.

4. Have a nucleus. 4. Often irregular in shape.

5. Often have chloroplasts

containing chlorophyll. 5. Do not contain plastids.

Complete question is;

A long, circular aluminum rod is attached at one end to a heated wall and transfers heat by convection to a cold fluid. If the diameter of the rod is doubled, by how much would the rate of heat removal change?

Answer:

182.84 %

Explanation:

Formula for rate of heat transfer of an infinite log fin is given as;

q_f1 = (π/2) × (hk)^(½)) × D^(3/2)) × (T_b - T_∞)

Where D is diameter.

Now, if the diameter of the rod is doubled, it means Diameter is now 2D.

Thus;

q_f2 = (π/2) × (hk)^(½)) × (2D^(3/2)) × (T_b - T_∞)

To find how much the rate of heat removal will change, we will calculate as follows;

((q_f2/q_f1) - 1) × 100

Plugging in the relevant expressions, we have;

([[(π/2) × (hk)^(½)) × ((2D)^(3/2)) × (T_b - T_∞)]/[(π/2) × (hk)^(½)) × (D^(3/2)) × (T_b - T_∞)]] - 1) × 100

Upon simplifying, we have;

(((2D)^(3/2))/(D^(3/2)) - 1) × 100

((2^(3/2)) - 1) × 100

This gives;

182.84 %

The question is incomplete. Here is the complete question.

Cars A nad B are racing each other along the same straight road in the following manner: Car A has a head start and is a distance [tex]D_{A}[/tex] beyond the starting line at t = 0. The starting line is at x = 0. Car A travels at a constant speed [tex]v_{A}[/tex]. Car B starts at the starting line but has a better engine than Car A and thus Car B travels at a constant speed [tex]v_{B}[/tex], which is greater than [tex]v_{A}[/tex].

Part A: How long after Car B started the race will Car B catch up with Car A? Express the time in terms of given quantities.

Part B: How far from Car B's starting line will the cars be when Car B passes Car A? Express your answer in terms of known quantities.

Answer: Part A: [tex]t=\frac{D_{A}}{v_{B}-v_{A}}[/tex]

              Part B: [tex]x_{B}=\frac{v_{B}D_{A}}{v_{B}-v_{A}}[/tex]

Explanation: First, let's write an equation of motion for each car.

Both cars travels with constant speed. So, they are an uniform rectilinear motion and its position equation is of the form:

[tex]x=x_{0}+vt[/tex]

where

[tex]x_{0}[/tex] is initial position

v is velocity

t is time

Car A started the race at a distance. So at t = 0, initial position is [tex]D_{A}[/tex].

The equation will be:

[tex]x_{A}=D_{A}+v_{A}t[/tex]

Car B started at the starting line. So, its equation is

[tex]x_{B}=v_{B}t[/tex]

Part A: When they meet, both car are at "the same position":

[tex]D_{A}+v_{A}t=v_{B}t[/tex]

[tex]v_{B}t-v_{A}t=D_{A}[/tex]

[tex]t(v_{B}-v_{A})=D_{A}[/tex]

[tex]t=\frac{D_{A}}{v_{B}-v_{A}}[/tex]

Car B meet with Car A after [tex]t=\frac{D_{A}}{v_{B}-v_{A}}[/tex] units of time.

Part B: With the meeting time, we can determine the position they will be:

[tex]x_{B}=v_{B}(\frac{D_{A}}{v_{B}-v_{A}} )[/tex]

[tex]x_{B}=\frac{v_{B}D_{A}}{v_{B}-v_{A}}[/tex]

Since Car B started at the starting line, the distance Car B will be when it passes Car A is [tex]x_{B}=\frac{v_{B}D_{A}}{v_{B}-v_{A}}[/tex] units of distance.

The distance traveled by the car A and car B  should be equal to the as they meet at the same position.

The time car B will catch the car A after is,

[tex]\dfrac{D_A}{v_B-v_A}[/tex]

The distance is the product of the speed of the body and the time taken to travel the distance.

Given information-

Car A has a head start and is a distance DA beyond the starting line at,

[tex]t=0[/tex]

Car A travels at a constant speed [tex]v_A[/tex].

Car B travels at a constant speed [tex]v_B[/tex].

The distance is the product of the speed of the body and the time taken to travel the distance.

The position equation from the motion for car A can be given as,

[tex]x_A=v_At+D_A[/tex]

The position equation from the motion for car B can be given as,

[tex]x_B=v_Bt[/tex]

The distance traveled by the car A and car B  should be equal to the as they meet at the same position. Thus,

[tex]x_A=x_B[/tex]

Put the values,

[tex]v_At+D_A=v_Bt\\v_At-v_Bt=-D_A\\t(v_B-v_A)=D_A\\t=\dfrac{D_A}{v_B-v_A}[/tex]

Hence the time car B will catch the car A after is,

[tex]\dfrac{D_A}{v_B-v_A}[/tex]

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The molecular mass of [tex]$O_{3}[/tex] is 0.43728kg

Molecular mass exists as a number equivalent to the totality of the atomic masses of the atoms in a molecule.  The totality of the atomic masses of all atoms in a molecule is established on a scale in which the atomic masses of hydrogen, carbon, nitrogen, and oxygen exist 1, 12, 14, and 16, respectively.

To compute the Molecular Mass of [tex]$O_{3}[/tex]

Atomic mass of oxygen(O) = 16

As [tex]$O_{3}[/tex] contains 3 atoms,

The molecular mass of [tex]$O_{3}[/tex]

= (16 x 3) = 48g/mol

Hence the mass of 9.11 moles O3

= 9.11 mol x 48g/mol

= 437.28g

= 0.43728kg

Therefore, the molecular mass of [tex]$O_{3}[/tex] is 0.43728kg

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A. Synthesis, decomposition, single-replacement, combustion

List of reaction types are redox reactions:

"Redox reactions are oxidation-reduction chemical reactions in which the reactants undergo a change in their oxidation states." The term 'redox' is a short form of reduction-oxidation. All the redox reactions can be broken down into two different processes – a reduction process and an oxidation process.

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Answer:

3.45e-11MV

that is ur answer

Answer:

Permittivity, also called electric permittivity, is a constant of proportionality that exists between electric displacement and electric field intensity.

Answer:

a periscope use total internal reflection to allow us to see things

the reflection happens at 45°

Explanation:

Answer:

326,000

Explanation:

One acre-foot equals about 326,000 gallons, or enough water to cover an acre of land, about the size of a football field, one foot deep. An average California household uses between one-half and one acre-foot of water per year for indoor and outdoor use.

Answer:

v = 79.3 km/h

Explanation:

      [tex]t_{1} = 161 min* \frac{60s}{1min} = 9660 s (1)[/tex]

      [tex]t_{2} = 23 min* \frac{60s}{1min} = 1380 s (2)[/tex]

      [tex]t_{3} = 211 min* \frac{60s}{1min} = 12660 s (3)[/tex]

      [tex]t_{tot} =t_{1} +t_{2} +t_{3} = 9660s + 1380s + 12660s = 23700s (4)[/tex]

       [tex]x_{1} = v_{1} * t_{1} (5)[/tex]

       [tex]x_{2} = v_{2} * t_{3} (6)[/tex]

       [tex]v_{1} = 107.0 km/h*\frac{1000m}{1km}*\frac{1h}{3600s} = 29.7 m/s (7)[/tex]

      [tex]v_{2} = 67.0 km/h*\frac{1000m}{1km}*\frac{1h}{3600s} = 18.6 m/s (8)[/tex]

       [tex]x_{1} = v_{1} * t_{1} = 29.7 m/s * 9660 s = 286902 m (9)[/tex]

       [tex]x_{2} = v_{2} * t_{3} = 18.6 m/s * 12660 s = 235476 m (10)[/tex]

       [tex]x_{tot} = x_{1} +x_{2} = 286902 m + 235476 m = 522378 m (11)[/tex]

       [tex]v_{avg} =\frac{\Delta x}{\Delta t} = \frac{522378m}{23700s} = 22.0 m/s (12)[/tex]

       [tex]v_{avg} = 22.0 m/s*\frac{1km}{1000m}*\frac{3600s}{1h} = 79.3 km/h (13)[/tex]

Answer:

a

Explanation:

Answer:

forcing in act

Explanation:

Answer:

letra A segundo o couculo a divisão e completa

Answer:

You need to know the accuracy to which you can read the ruler:

Suppose that you can read the read the ruler to the nearest milimeter

A = L * W     your calculated area of the rectangle

A + ΔA = (L + ΔL) * (W + ΔW) = L W + L ΔW + W * ΔL + ΔL ΔA

Or ΔA =  L ΔW + W ΔL

Where we have subtracted A = L * W and the term ΔL * ΔA is very small

So (5 + .1) * (2 + .1) - 5 * 2 = .1 * 2 + .1 * 5 = .7 cm^2

Then you report A = 10 cm^2 +- .7 cm^2    including the - sign for completeness

Answer:

0.248

Explanation:

Initial speed u = 13.6

Final speed v = 0

Distance s = 38.1

We have umg = ma

We make u subject of the formula

u = a/g

V² = u² + 2as

a = v²-u²/2s

We substitute the values into the above

a = 0-(13.6)²/2*38.1

a = 184.96/76.2

a = 2.427m/sec

Remember that

u = a/g

u = 2.427/9.8

= 0.2476

This is approximately 0.248

This is the minimum coefficient of friction required to keep the crate from sliding.

Answer:

a) 5.09 seconds

b) 107.07 meters

Explanation:

a) As we know

[tex]t_2- t_1 = \sqrt{\frac{2 X}{a} }[/tex]

Substituting the given values we get

[tex]t_2 - t_1 = \sqrt{\frac{2 * 52}{4} } \\t_2 - t_1 = 5.09[/tex]

It takes 5 .09 s for the motorcycle to accelerate until it catches up with the car

b)

[tex]X_{t`2} = v_i \sqrt{\frac{2X}{a} } + 0.5 a\sqrt{\frac{2X}{a} }\\X_{t`2} = (v_i + 0.5 a) \sqrt{\frac{2X}{a} }\\X_{t`2} = ( 19 + 2) \sqrt{\frac{2* 52}{4} }\\X_{t`2} = 21 * 5.09\\X_{t`2} = 107.07[/tex]

Answer:

Earth's atmosphere traps energy from the sun. Which is a direct result of the trapping of energy by earth's atmosphere? Earth has moderate temperatures.

Explanation: